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\title{D\'emarche de base en s\'eries temporelles (compl\'ements du Chapitre 1)}
\author{Yves Aragon\footnote{aragon@cict.fr} \cr
{\normalsize Universit\'e Toulouse 1 Capitole} }
\begin{document}
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\begin{Exercice}[Op\'erateur retard]
Calculer : (a) $(1 - 0.9\: \bm)1$,  (b) $(1 - 0.9\: \bm)t$, (c) $\frac{1}{1 - 0.9 \:\bm} t$, o\`u 1 d\'esigne la fonction constante, et $t$ la fonction  $ t \mapsto t \;\forall t$.
\end{Exercice}

\textbf{R\'eponse}
\begin{itemize}
\item
(a) Op\'erateur retard appliqu\'e \`a  la fonction constante $t  \mapsto 1 \;\forall t$,
\[(1 - 0.9\: \bm)1 = 1 - 0.9 \:\bm 1 = 1 - 0.9 = 0.1\]
et
\[
(1 - \bm) 1 = 0,
\]
tout simplement, l'accroissement d'une fonction constante est nul.
\item 
(b) Op\'erateur retard appliqu\'e \`a  la fonction $ t \mapsto t \;\forall t$
\[(1 - 0.9\: \bm)t = t - 0.9\: \bm t = t - 0.9 (t-1) = 0.1 t +0.9\]
et
\[
(1 - \bm) t = t - (t-1) = 1.
\]
\item (c) Fraction rationnelle de l'op\'erateur retard :
\[
\frac{1}{1 - 0.9\: \bm} 1 = (1 +  0.9 \:\bm  +  (0.9)^2 \:\bm^2   +\cdots)1  = 1+ 0.9 + (0.9)^2+ \cdots = \frac{1}{1 - 0.9} = 10,
\]
\begin{multline}
\frac{1}{1 - 0.9\: \bm} t = (1 +  0.9 \:\bm  +  0.9^2 \:\bm^2   +\cdots ) t = t+ 0.9 (t-1) + 0.9^2 (t-2) + \cdots = \\
t \frac{1}{1 - 0.9} -0.9 \frac{1}{(1-0.9)^2} = \cdots=10\: t - 90.
\end{multline}
\end{itemize}

\begin{Exercice} [SARMA et op\'erateur retard]
Ecrire
le mod\`ele SARMA (\ref{sarma.sim}) \`a l'aide de l'op\'erateur retard.
\begin{eqnarray} \label{sarma.sim}
y_t - 50 = 0.9 (y_{t-12}-50) + z_t - 0.7 z_{t-1},
\end{eqnarray}
o\`u les $z_t$ sont i.i.d. $\nor(0,17.5)$.

\end{Exercice}
\textbf{R\'eponse.}
Mod\`ele SARMA (\ref{sarma.sim}) exprim\'e \`a l'aide de l'op\'erateur retard :
\begin{subequations} \label{E:gp}
  \begin{gather}
   y_t - 50 =  0.9 \;(\bm^{12} y_t-50) + z_t - 0.7\; \bm z_t =  0.9\; \bm^{12}( y_t-50) + (1 - 0.7 \; \bm) z_t\label{E:gp1} \\
   ( 1 - 0.9 \; \bm^{12} ) (y_t - 50)  = (1 - 0.7 \;\bm) z_t \label{E:gp2} \\
   y_t - 50  = \frac{ 1 - 0.7\; \bm}{1 - 0.9 \;\bm^{12}} z_t \label{E:gp3}\\
   y_t   = 50 + \frac{ 1 - 0.7\; \bm}{1 - 0.9 \;\bm^{12}} z_t.  \label{E:gp4}
  \end{gather}
\end{subequations}
Alternativement, comme $( 1 - 0.9 \; \bm^{12} ) 50 = 0.1\times 50 = 5$,  en multipliant les deux c\^ot\'es de
(\ref{E:gp4}) par $1 - 0.9 \;\bm^{12} $, on obtient :
\begin{eqnarray} \label{Egp}
 ( 1 - 0.9 \; \bm^{12} ) y_t    = 5 + (1 - 0.7 \;\bm) z_t.
 \end{eqnarray}
Pour obtenir ces diff\'erentes \'ecritures du SARMA (\ref{sarma.sim}) nous avons exploit\'e le fait, not\'e plus haut, que l'op\'erateur
retard appliqu\'e \`a une constante donne cette constante.
Chacune de ces \'ecritures a ses m\'erites et il est important de savoir passer de l'une \`a l'autre en fonction notamment
de l'interlocuteur et des outils informatiques qui privil\'egient l'une ou l'autre.

\end{document}



